A 276.0 g piece of granite, heated to 596.0°C in a campfire, is dropped into 1.45 L water (d = 1.00 g/mL) at 25.0°C.

The molar heat capacity of water is cp, water = 75.3 J / (mol ·°C), and the specific heat of granite is cs, granite = 0.790 J / (g ·°C).

Calculate the final temperature of the granite in Celsius.

Answer;

Final temperature of the granite ≈ 44.82 °C

Explanation;

Let T be the equilibrium temperature (in °C) to be found.

Moles of water;

(1.45 L H2O) x (1000 mL/L) x (1.00 g/mL) / (18.01 g H2O/mol)

= 80.512 mol H2O

Heat gained by water;

= (75.3 J / (mol ·°C)) x (80.512 mol) x (T - 25.0) °C

= (6062.465 T - 151561.6)

Heat lost by granite;

= (0.790 J / (g ·°C)) x (276 g) x (596 - T) °C

= (129951.84 - 218.04 T)

Set the two expressions for heat gained/lost equal to each other:

6062.465 T - 151561.6 = 129951.84 - 218.04 T

6280.505 T = 281513.44

T = 44.8234

T ≈ 44.82 °C